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144.二叉树的前序遍历.ts
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144.二叉树的前序遍历.ts
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/*
* @lc app=leetcode.cn id=144 lang=typescript
*
* [144] 二叉树的前序遍历
*
* https://leetcode.cn/problems/binary-tree-preorder-traversal/description/
*
* algorithms
* Easy (71.30%)
* Likes: 1050
* Dislikes: 0
* Total Accepted: 839.2K
* Total Submissions: 1.2M
* Testcase Example: '[1,null,2,3]'
*
* 给你二叉树的根节点 root ,返回它节点值的 前序 遍历。
*
*
*
* 示例 1:
*
*
* 输入:root = [1,null,2,3]
* 输出:[1,2,3]
*
*
* 示例 2:
*
*
* 输入:root = []
* 输出:[]
*
*
* 示例 3:
*
*
* 输入:root = [1]
* 输出:[1]
*
*
* 示例 4:
*
*
* 输入:root = [1,2]
* 输出:[1,2]
*
*
* 示例 5:
*
*
* 输入:root = [1,null,2]
* 输出:[1,2]
*
*
*
*
* 提示:
*
*
* 树中节点数目在范围 [0, 100] 内
* -100
*
*
*
*
* 进阶:递归算法很简单,你可以通过迭代算法完成吗?
*
*/
interface TreeNode {
val: number;
left: TreeNode | null;
right: TreeNode | null;
}
// @lc code=start
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function preorderTraversal(root: TreeNode | null): number[] {
const ans: number[] = [];
traversal(root);
function traversal(node: TreeNode | null) {
if (!node) return;
ans.push(node.val);
traversal(node.left);
traversal(node.right);
}
return ans;
}
// @lc code=end